I came up with a proof but I am not sure if that is correct.
I am not sure whether this is rigorous proof, but I think I have a proof for the fact that $[(1 + \sqrt{3})^{2n+1}] = k2^{n+1}$ for $k \in N$.
It is primarily based on the idea that $2 = (\sqrt{3} - 1) (\sqrt{3} + 1)$
So, it goes something like this:
In order to prove that, [$(1 + \sqrt{3})^{2n+1}] = k2^{n+1}$
We need to prove that,
$k2^{n+1} \le (1 + \sqrt{3})^{2n+1} < k2^{n+1} + 1$
Which same as proving,
$k((\sqrt(3) - 1)(\sqrt(3) + 1))^{n+1} \le (1 + \sqrt{3})^{2n+1} < k((\sqrt(3) - 1)(\sqrt(3) + 1))^{n+1} + 1$
Which is same as proving,
$k(\sqrt(3) - 1)^{n+1} \le (1 + \sqrt{3})^{n} < k(\sqrt(3) - 1)^{n+1} + \frac{1}{(1 + \sqrt{3})^{n + 1}}$
Which is same as proving,
$[(1 + \sqrt{3})^{n}] = [k(\sqrt(3) - 1)^{n+1}]$ or $[(1 + \sqrt{3})^{n}] = [k(\sqrt(3) - 1)^{n+1}] + 1$
Because $\frac{1}{(1 + \sqrt{3})^{n + 1}} < 1$ as $n \ge 0$
What does proving $[(1 + \sqrt{3})^{n}] = [k(\sqrt(3) - 1)^{n+1}]$ really mean ?
I think it means (and I may be wrong) that we can always find such a $k$ for any given $n$
From the equation $[(1 + \sqrt{3})^{n}] = [k(\sqrt(3) - 1)^{n+1}]$ , it looks very obvious that such a k should always exist.
Just to add more rigor let’s assume that,
$[(1 + \sqrt{3})^{n}] = p$
where $p \in N$
So this reduces to finding a $k$ such that $[k(\sqrt(3) - 1)^{n+1}] = p$
which means $p \le k(\sqrt(3) - 1)^{n+1} < p + 1$
Consider the first half of the inequality
$p \le k(\sqrt(3) - 1)^{n+1} $
$\frac{p}{(\sqrt(3) - 1)^{n+1}} \le k$
$\frac{p(\sqrt(3) + 1)^{n + 1}}{2^{n+1}} \le k$ $.... (1)$
Clearly for any given natural number $p$ we can always find such a positive integer $k$
Let’s consider the other half of the inequality,
$k(\sqrt(3) - 1)^{n+1} < p + 1$
$k < \frac{p + 1}{(\sqrt(3) - 1)^{n+1}}$
$k < \frac{(p + 1)(\sqrt(3) + 1)^{n + 1}}{2^{n+1}}$ $.... (2)$
Clearly a positive integer satisfying both (1) and (2) will always exist because,
$\frac{(p + 1)(\sqrt(3) + 1)^{n}}{2^{n+1}} - \frac{(p)(\sqrt(3) + 1)^{n}}{2^{n+1}} = \frac{(\sqrt(3) + 1)^{n + 1}}{2^{n+1}}( p + 1 - p) = \frac{(\sqrt(3) + 1)^{n + 1}}{2^{n+1}}$
and $\frac{(\sqrt(3) + 1)^{n + 1}}{2^{n+1}} > 1$ and hence $k$ always exists.
What does this prove ?? Well since every step in the above derivation is reversible I guess we can conclude that $[(1 + \sqrt{3})^{2n+1}] $is divisible by $2^{n+1}$
Similarly we can also prove that $[(1 + \sqrt{3})^{n}] = [k(\sqrt(3) - 1)^{n+1}] + 1$ will also always have a solution, by assuming $[(1 + \sqrt{3})^{n}] - 1 = p$
P.S : I will be thankful if someone reviews my proof :)
